Ask Singapore Homework?
Upload a photo of a Singapore homework and someone will email you the solution for free.
Question
secondary 3 | A Maths
4 Answers Below
Anyone can contribute an answer, even non-tutors.
How to do qn 4(a) and 4(b)
Date Posted: 4 years ago
Views: 422
Question 4b has a fault, as I suspected. Try k = 1 and k = 4, the line does touch the curve.
See 4 Answers
done
{{ upvoteCount }} Upvotes
clear
{{ downvoteCount * -1 }} Downvotes
First one. Now stuck at the second. I will need to inspect my answer for the second one first.
Date Posted:
4 years ago
done
{{ upvoteCount }} Upvotes
clear
{{ downvoteCount * -1 }} Downvotes
I thought the answer should be this. By restricting the values of k, we will see the line does not intersect the curve for real values of x. And we should prove b²-4ac <0 and not the other way?
Date Posted:
4 years ago
It’s more of a question phrasing issue. Yes, for k between 1 and 4, the curve will not touch the x-axis for all x.
But the question suggests that we need to show that the curve is positive regardless of the constant k used.
But the question suggests that we need to show that the curve is positive regardless of the constant k used.
A good question will not have such ambiguities and should sound like this.
Show that for some values of k to be determined, the curve...
Show that for some values of k to be determined, the curve...
Ok noted. Thanks a lot.
done
{{ upvoteCount }} Upvotes
clear
{{ downvoteCount * -1 }} Downvotes
First question mistake, it should be ‘for all real values of k’, not ‘for all real values of x’, since x and y are the variables to be plotted on the graph.
Second mistake, when I used b2 - 4ac, I obtained 4k2 - 20k + 16 which can be factorised nicely into 4 (k - 4) (k - 1); the fact that it can be factorised means that there are indeed values of k for which b2 - 4ac = 0 and for which b2 - 4ac > 0.
In light of these two, there is indeed a question fault.
Just out of curiosity, which paper is this taken from?
Second mistake, when I used b2 - 4ac, I obtained 4k2 - 20k + 16 which can be factorised nicely into 4 (k - 4) (k - 1); the fact that it can be factorised means that there are indeed values of k for which b2 - 4ac = 0 and for which b2 - 4ac > 0.
In light of these two, there is indeed a question fault.
Just out of curiosity, which paper is this taken from?
Date Posted:
4 years ago
Unless k and y are to be plotted.
y = 4x + k
y + x2 = 2kx
If k is indeed a variable (and x can be set to any number), then the two equations reduces to a linear equation y = mk + c. For many values of x, the lines will intersect. So again, either way, the question is flawed.
y = 4x + k
y + x2 = 2kx
If k is indeed a variable (and x can be set to any number), then the two equations reduces to a linear equation y = mk + c. For many values of x, the lines will intersect. So again, either way, the question is flawed.
Chij sngs ca1
done
{{ upvoteCount }} Upvotes
clear
{{ downvoteCount * -1 }} Downvotes
Proof of question fault
Date Posted:
4 years ago
360 Tutoring Program